Key to Quiz #8 - Friday, May 24


ODD (i.e.top) PROBLEMS:

  1. Calculate the pH of a 0.03 M nitric acid solution. Is it...
    1. 12.48
    2. 1.52
    3. 5.48
    4. 3.00
    5. 0.30
    ANSWER: Nitric acid is a strong acid. The [H3O-] is therefore equal to 0.03M and the pH is simple the negative log of this which equals 1.52, B.

  2. The pH of a strong base solution is 13.60. Calculate the concentration of strong base. Is it...
    1. 0.120 M
    2. 13.60 M
    3. 2.5 x 10-14 M
    4. 0.20 M
    5. 0.398 M
    ANSWER: The concentration of strong base will equal [OH-]. One therefore obtains [H3O-] from the pH, and divides this number into Kw to obtain [OH-]. The solution is 0.398 M, E.

  3. Calculate the pH of a 0.16 M solution of acetic acid. The Ka is 1.8 x 10-5. Is it...
    1. 4.75
    2. 1.77
    3. 2.77
    4. 9.43 x 10-6
    5. 3.75
    ANSWER: The expedient method for solving for the pH works fine here because the acid conentration is substantial. One obtains [H3O-] by taking the square root of the product of Ka with [A.A.]. Then the negative log yields the pH, which is 2.77 in this case, C.

  4. The pH of a weak acid solution is 2.89. The concnetration of weak acid is 0.100 M. Calculate Ka. Is it.
    1. 1.28 x 10-3
    2. 1.66 x 10-4
    3. 1.66 x 10-6
    4. 4.78
    5. 1.66 x 10-5
    ANSWER: Again, because the weak acid concentration is relatively high, Ka can be obtained by squaring the [H3O-] and dividing by the concentration of weak acid. [H3O-] is obtained in the usual way from the pH. The result in this case is 1.66 x 10-5, E.


EVEN (i.e. bottom) PROBLEMS:

  1. Calculate the pH of a 0.02 M KOH solution (KOH is a strong base). Is it...
    1. 1.70
    2. 12.00
    3. 12.30
    4. 7.30
    5. 1.73
    ANSWER: Because KOH is a strong base, the [OH-] will equal 0.02M. The [H3O-] is then obtained by dividing Kw by 0.02, and the log of this result yields the pH, 12.30, C.

  2. The pH of a strong acid solution is 0.82. Calculate the concentration of strong acid. Is it...
    1. 0.15 M
    2. 4.67 M
    3. 13.2 M
    4. 6.61 M
    5. 0.82 M
    ANSWER: The concentration of strong acid will equal [H3O-]. In this case, [H3O-] = 0.151 M, A.

  3. Calculate the pH of a 0.091 M solution of HCN. The Ka is 2.1 x 10-9. Is it...
    1. 1.47
    2. 4.86.
    3. 3.27
    4. 1.38 x 10-5
    5. 2.21
    ANSWER: The expedient method for solving for the pH works fine here because the acid concentration is substantial. One obtains [H3O-] by taking the square root of the product of Ka with [A.A.]. Then the negative log yields the pH, which is 4.86 in this case, B.

  4. The pH of a weak acid solution is 4.11. The concentration of weak acid is 0.100 M. Calculate Ka. Is it.
    1. 7.76 x 10-5
    2. 6.02 x 10-8
    3. 1.02 x 10-8
    4. 6.02 x 10-9
    5. 6.02 x 10-7
    ANSWER: Again, because the weak acid concentration is relatively high, Ka can be obtained by squaring the [H3O-] and dividing by the concentration of weak acid. [H3O-] is obtained in the usual way from the pH. The result in this case is 6.02 x 10-8, B.
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Last modified: May 24, 1996